In: Statistics and Probability
A recent study found that 64 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 36 grams of potato chips as compared to a mean of 26 grams for 54 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the celebrity-endorsed commercial was 21.1 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.9 grams. Complete parts (a) through (c) below.
a. Assuming that the population variances are equal and alphaequals0.05, is there evidence that the mean amount of potato chips eaten was significantly higher for the children who watched the celebrity-endorsed commercial? Let population 1 be the weights of potato chips eaten by children who watched the celebrity-endorsed commercial and let population 2 be the weights of potato chips eaten by children who watched the alternative food snack commercial. What are the correct null and alternative hypotheses? a)What is the test statistic? (Round to two decimal places)
what is the p value? (round to three decimal places)
b) Assuming that the population variances are equal, construct a 95% confidence interval estimate of the difference mu 1 minus mu 2 between the mean amount of potato chips eaten by the children who watched the celebrity-endorsed commercial and children who watched the alternative food snack commercial.determine the 95% confidence interval using technology, rounding to two decimal places.
c) Compare and discuss the results of (a) and (b). Check to make sure the results of (a) and (b) can be compared. Recall that confidence intervals are comparable only to the results of a two-tail hypothesis test. Since the test done in part (a) is a one-tail hypothesis test, these two results cannot be meaningfully compared.
a.
Given that,
mean(x)=36
standard deviation , s.d1=21.1
number(n1)=64
y(mean)=26
standard deviation, s.d2 =129
number(n2)=54
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.66
since our test is right-tailed
reject Ho, if to > 1.66
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2
)/(n1+n2-2)
s^2 = (63*445.21 + 53*16641) / (118- 2 )
s^2 = 7845.0106
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=36-26/sqrt((7845.0106( 1 /64+ 1/54 ))
to=10/16.3663
to=0.611
| to | =0.611
critical value
the value of |t α| with (n1+n2-2) i.e 116 d.f is 1.66
we got |to| = 0.611 & | t α | = 1.66
make decision
hence value of |to | < | t α | and here we do not reject
Ho
p-value: right tail -ha : ( p > 0.611 ) = 0.27119
hence value of p0.05 < 0.27119,here we do not reject Ho
ANSWERS
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i.
null, Ho: u1 = u2
alternate, H1: u1 > u2
ii.
test statistic: 0.611
critical value: 1.66
decision: do not reject Ho
iii.
p-value: 0.27119
we do not have enough evidence to support the claim that he mean
amount of potato chips eaten was significantly higher for the
children who watched the celebrity-endorsed commercial
b.
TRADITIONAL METHOD
given that,
mean(x)=36
standard deviation , s.d1=21.1
number(n1)=64
y(mean)=26
standard deviation, s.d2 =129
number(n2)=54
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2
)/(n1+n2-2)
s^2 = (63*445.21 + 53*16641) / (118- 2 )
s^2 = 7845.011
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 7845.011 * (1/64+1/54) )
=16.366
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and value of |t α| with
(n1+n2-2) i.e 116 d.f is 1.981
margin of error = 1.981 * 16.366
= 32.422
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (36-26) ± 32.422 ]
= [-22.422 , 42.422]
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DIRECT METHOD
given that,
mean(x)=36
standard deviation , s.d1=21.1
sample size, n1=64
y(mean)=26
standard deviation, s.d2 =129
sample size,n2 =54
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 36-26) ± t a/2 * sqrt( 7845.011 * (1/64+1/54) ]
= [ (10) ± 32.422 ]
= [-22.422 , 42.422]
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interpretations:
1. we are 95% sure that the interval [-22.422 , 42.422]contains the
true population proportion
2. If a large number of samples are collected, and a confidence
interval is created
for each sample, 95% of these intervals will contains the true
population proportion
c.
ANSWERS:
a.
we do not have enough evidence to support the claim that he mean
amount of potato chips eaten was significantly higher for the
children who watched the celebrity-endorsed commercial
b.
95% sure that the interval [-22.422 , 42.422]
confidence intervals are comparable only to the results of a
two-tail hypothesis test.
part (a) is one tailed test.